Question

If Punxsutawney Phil sees his shadow on Feb. 2, six more weeks of winter weather lay ahead; no shadow indicates an early spring. Phil, a groundhog, has been forecasting the weather on Groundhog Day for more than 120 years, but just how good is he at his job?

Hypothesis tests: Does Punxsutawney Phil predict the weather accurately? I believe Punxsutawney Phil’s accurate predictions will be less than 50%. From a sample of 100 data values we find that Phil predicted the weather accurately 36 times. At the 0.05 level of significance, can we conclude that Phil will predict the weather less than 50% of the time?

H₀:

H_a:

α=

Decision Rule:

Critical Value = ___________

Standardized Test Statistic = p - p0 /p(1-p)/n =.

At the _________ level of significance there __________ reason to believe ___________________

__________________________________________________________________

Confidence Interval: Construct a 95% confidence interval about the true portion.

We are _____ confident that Punxsutawney Phil will predict the weather between

__________ and ___________ percent of the time.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.5
Alternative Hypothesis, Ha: p < 0.5

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.64.
Hence reject H0 if z < -1.64

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.36 - 0.5)/sqrt(0.5*(1-0.5)/100)
z = -2.80

reject the null hypothesis.

At the 0.05 level of significance there is reason to believe that Phil will predict the weather less than 50% of the time

sample proportion, = 0.36
sample size, n = 100
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.36 * (1 - 0.36)/100) = 0.048

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.048
ME = 0.0941

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.36 - 1.96 * 0.048 , 0.36 + 1.96 * 0.048)
CI = (0.2659 , 0.4541)

We are __95%___ confident that Punxsutawney Phil will predict the weather between

0.2659 = 26.59% and 0.4541 = 45.41% percent of the time.